行列の計算例題（行列式（余因子展開））

　　　　　　　　　　


例題：次の行列式を余因子展開を使って計算せよ。

\[ \left|\,\begin{array}{@{}rwr{20pt}wr{20pt}wr{20pt}@{}}1 & {-}1 & {-}1 & 0 \\ {-}2 & 1 & 2 & {-}1 \\ {-}1 & {-}2 & 2 & {-}2 \\ 2 & {-}2 & {-}2 & {-}2\end{array}\,\right| \]

解答

\( \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}@{}}1 & {-}1 & {-}1 & 0 \\ {-}2 & 1 & 2 & {-}1 \\ {-}1 & {-}2 & 2 & {-}2 \\ 2 & {-}2 & {-}2 & {-}2\end{array}\,\right| \qquad \) （余因子展開） 第4列で展開する : [0, -1, -2, -2]

\( \qquad\quad = (-1)^{4-2}\times(-1)\times\left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}1 & {-}1 & {-}1 \\ {-}1 & {-}2 & 2 \\ 2 & {-}2 & {-}2\end{array}\,\right| + (-1)^{4-3}\times(-2)\times\left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}1 & {-}1 & {-}1 \\ {-}2 & 1 & 2 \\ 2 & {-}2 & {-}2\end{array}\,\right| + (-1)^{4-4}\times(-2)\times\left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}1 & {-}1 & {-}1 \\ {-}2 & 1 & 2 \\ {-}1 & {-}2 & 2\end{array}\,\right| \)

\( \qquad\quad = (-1)^{4-2}\times(-1)\times0 + (-1)^{4-3}\times(-2)\times0 + (-1)^{4-4}\times(-2)\times(-1) \)

\( \qquad\quad = 2 \)

\( \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}1 & {-}1 & {-}1 \\ {-}1 & {-}2 & 2 \\ 2 & {-}2 & {-}2\end{array}\,\right| \qquad \) （余因子展開） 第3列で展開する : [-1, 2, -2]

\( \qquad\quad = (-1)^{3-1}\times(-1)\times\left|\,\begin{array}{@{}rwr{15pt}@{}}{-}1 & {-}2 \\ 2 & {-}2\end{array}\,\right| + (-1)^{3-2}\times2\times\left|\,\begin{array}{@{}rwr{15pt}@{}}1 & {-}1 \\ 2 & {-}2\end{array}\,\right| + (-1)^{3-3}\times(-2)\times\left|\,\begin{array}{@{}rwr{15pt}@{}}1 & {-}1 \\ {-}1 & {-}2\end{array}\,\right| \)

\( \qquad\quad = (-1)^{3-1}\times(-1)\times6 + (-1)^{3-2}\times2\times0 + (-1)^{3-3}\times(-2)\times(-3) \)

\( \qquad\quad = 0 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}{-}1 & {-}2 \\ 2 & {-}2\end{array}\,\right| = (-1)\times(-2) - (-2)\times2 = 6 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}1 & {-}1 \\ 2 & {-}2\end{array}\,\right| = 1\times(-2) - (-1)\times2 = 0 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}1 & {-}1 \\ {-}1 & {-}2\end{array}\,\right| = 1\times(-2) - (-1)\times(-1) = -3 \)

\( \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}1 & {-}1 & {-}1 \\ {-}2 & 1 & 2 \\ 2 & {-}2 & {-}2\end{array}\,\right| \qquad \) （余因子展開） 第3列で展開する : [-1, 2, -2]

\( \qquad\quad = (-1)^{3-1}\times(-1)\times\left|\,\begin{array}{@{}rwr{15pt}@{}}{-}2 & 1 \\ 2 & {-}2\end{array}\,\right| + (-1)^{3-2}\times2\times\left|\,\begin{array}{@{}rwr{15pt}@{}}1 & {-}1 \\ 2 & {-}2\end{array}\,\right| + (-1)^{3-3}\times(-2)\times\left|\,\begin{array}{@{}rwr{15pt}@{}}1 & {-}1 \\ {-}2 & 1\end{array}\,\right| \)

\( \qquad\quad = (-1)^{3-1}\times(-1)\times2 + (-1)^{3-2}\times2\times0 + (-1)^{3-3}\times(-2)\times(-1) \)

\( \qquad\quad = 0 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}{-}2 & 1 \\ 2 & {-}2\end{array}\,\right| = (-2)\times(-2) - 1\times2 = 2 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}1 & {-}1 \\ 2 & {-}2\end{array}\,\right| = 1\times(-2) - (-1)\times2 = 0 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}1 & {-}1 \\ {-}2 & 1\end{array}\,\right| = 1\times1 - (-1)\times(-2) = -1 \)

\( \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}1 & {-}1 & {-}1 \\ {-}2 & 1 & 2 \\ {-}1 & {-}2 & 2\end{array}\,\right| \qquad \) （余因子展開） 第3列で展開する : [-1, 2, 2]

\( \qquad\quad = (-1)^{3-1}\times(-1)\times\left|\,\begin{array}{@{}rwr{15pt}@{}}{-}2 & 1 \\ {-}1 & {-}2\end{array}\,\right| + (-1)^{3-2}\times2\times\left|\,\begin{array}{@{}rwr{15pt}@{}}1 & {-}1 \\ {-}1 & {-}2\end{array}\,\right| + (-1)^{3-3}\times2\times\left|\,\begin{array}{@{}rwr{15pt}@{}}1 & {-}1 \\ {-}2 & 1\end{array}\,\right| \)

\( \qquad\quad = (-1)^{3-1}\times(-1)\times5 + (-1)^{3-2}\times2\times(-3) + (-1)^{3-3}\times2\times(-1) \)

\( \qquad\quad = -1 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}{-}2 & 1 \\ {-}1 & {-}2\end{array}\,\right| = (-2)\times(-2) - 1\times(-1) = 5 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}1 & {-}1 \\ {-}1 & {-}2\end{array}\,\right| = 1\times(-2) - (-1)\times(-1) = -3 \)

\( \qquad \left|\,\begin{array}{@{}rwr{15pt}@{}}1 & {-}1 \\ {-}2 & 1\end{array}\,\right| = 1\times1 - (-1)\times(-2) = -1 \)